Find the equations of the line passing through the point `(3,0,1)` parallel to the planes `x+2y=0` and `3y-z=0`.

Explanation

Equation of the two planes are `x+2y=0` and `3y-z=0`
Let `vecn_(1)` and `vecn_(2)` are the normals to the two planes respectively.
`:. vecn_(1)=hati+2hatj` and `vecn_(2)=3hati-hatk`
Therefore, `vecb=vecn_(1)xxvecn_(2)=|(hati, hatj, hatk),(1,2,0),(0,3,-1)|`
`=hati(-2)-hatj(-1)+hatk(3)`
`=-2hati+hatj+3hatk`
So the equation of the lines through the point `(3,0,1)` and parallel to the given two plane
`(x-3)hati+(y-0)hatj+(z-1)hatk+lamda(-2hatk+hatj+3hatk)`
`implies (x-3)hati+yhatj+(z-1)hatk+lamda(-2hati+hatj+3hatk)`